Magnetic field by ideal toroidal solenoid

In summary, the magnetic field generated by an ideal toroidal solenoid can be calculated using the integral of the Biot-Savart law.
  • #1
DavideGenoa
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5
I am trying to calculate the magnetic field generated by an ideal toroidal solenoid by using the integral of the Biot-Savart law. I do not intend to use Ampère's circuital law.
Let ##I## be the intensity of the current flowing in each of the ##N## loops of the solenoid, which I will consider an ideal continuous solenoid from this point.
If ##\mathbf{r}(u,v):[0,2\pi]^2\to\mathbb{R}^3##, ##\mathbf{r}(u,v)=(b+a\cos v)\cos u\mathbf\,{i}+(b+a\cos v)\sin u\,\mathbf{j}+a\sin v\mathbf\,{k}## is a parametrization of the torus, I would say that, in an "infinitesimal spire" of the ideal solenoid, generated by the rotation of ##du## radians of the circumference generating the torus, an "infinitesimal current" ##\frac{IN}{2\pi}du## flows and therefore I would think that the magnetic field at ##\mathbf{x}## could be expressed by $$\frac{\mu_0}{4\pi} \int_{0}^{2\pi}\int_0^{2\pi}\frac{IN \,\partial_v\mathbf{r}(u,v) \times(\mathbf{x}-\mathbf{r}(u,v) )}{2\pi\|\mathbf{x}-\mathbf{r}(u,v)\|^3} dudv.$$
Am I right?
I thank anybody for any answer.
 
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  • #2
I think everything looks good, but maybe it's simpler to use the (Coulomb gauge) vector potential, which is given as
$$\vec{A}(\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|},$$
and then the magnetic field by
$$\vec{B} = \vec{\nabla} \times \vec{A}.$$
Note, however that in this case, due to the symmetry of the field, the use of Ampere's Law in integral form is much simpler here.
 
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  • #3
Thank you so much for your answer! By explicitly writing the formula, I find that, for a point ##\mathbf{x}=x\mathbf{i}+z\mathbf{k}##, the ##x## and ##z## components of the field ##\mathbf{B}## are zero because of reasons of symmetry and $$\mathbf{B}(x,0,z)=\frac{\mu_0 IN}{8\pi^2}\int_0^{2\pi}\int_0^{2\pi}\frac{ax\cos v-ab\cos u\cos v-a^2\cos u+az\cos u\sin v}{(x^2+b^2+a^2+z^2-2x(b+a\cos v)\cos u+2ab\cos v-2az\sin v)^{3/2}}dudv\mathbf{k}$$which, if it agrees with the result given by my books for an ideal solenoid, should be zero at the exterior of the torus and ##-\frac{\mu_0 NI}{2\pi x}\mathbf{j}## for ##z=0##, ##|x|\in (b-a,b+a)## where ##a## is the radius of the circular section of the torus.
Are my calculation correct until now? If they are, I find the integral quite difficult... Has anybody any idea of how to calculate it?
I ##\infty##-ly thank you all!
 

Related to Magnetic field by ideal toroidal solenoid

1. What is a toroidal solenoid?

A toroidal solenoid is a type of electromagnet that is shaped like a doughnut, with a hollow center. It is made up of a tightly wound coil of wire that produces a strong magnetic field when an electric current is passed through it.

2. How does a toroidal solenoid produce a magnetic field?

As an electric current flows through the wire coil of a toroidal solenoid, it creates a circular magnetic field around the coil. This magnetic field is amplified by the shape of the solenoid, resulting in a strong and uniform magnetic field inside the hollow center of the doughnut shape.

3. What are the properties of the magnetic field produced by a toroidal solenoid?

The magnetic field produced by a toroidal solenoid is strong, uniform, and tightly confined within the hollow center of the doughnut shape. It is also highly concentrated and can be directed and controlled by adjusting the current flowing through the wire coil.

4. What are the real-world applications of toroidal solenoids?

Toroidal solenoids have a wide range of applications in various industries, including medical equipment, particle accelerators, and magnetic storage devices. They are also commonly used in electromechanical devices such as relays and actuators.

5. Are there any limitations to the magnetic field produced by a toroidal solenoid?

While toroidal solenoids can produce strong and uniform magnetic fields, they are limited in terms of the size and shape of the field they can produce. The magnetic field can also weaken as distance from the solenoid increases. Additionally, the strength of the field is dependent on the current flowing through the coil, which may be limited by the capacity of the power source.

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